\documentclass{ctexart}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{geometry}
\geometry{a4paper,left=1.5cm,right=1.5cm,top=3cm,bottom=2.5cm}
\usepackage{amssymb}
\usepackage{amsthm}
\newtheorem{theorem}{定理}[section]  % 按 section 编号
\newtheorem{lemma}{引理}
\newtheorem{remark}{注}
\newtheorem{Proof}{证明：}
\newtheorem{introduction}{引言}
\title{实对称矩必可在实数域上相似对角化}
\author{陈冠宇 3200102033}
\date{\today}
\begin{document}

\maketitle
\begin{introduction}
  实对称矩必可在实数域上相似对角化.这个结论在线性代数中十分重要。
\end{introduction}
\section{问题描述}

\begin{theorem}
  实对称矩必可在实数域上相似对角化.即$\forall A \in \mathbb{R}^n,\exists $正交矩阵$T,s.t.A=T^{-1}\begin{pmatrix}
    \lambda_1 &   &   &   \\
    & \lambda_2 &   &   \\
    &   & \ddots  &   \\
    &   &   & \lambda_{n}
  \end{pmatrix} T$
\end{theorem}
\subsection*{证明}
\begin{lemma}
  实对称矩阵的特征值是实数。
\end{lemma}
\begin{proof}
  设X是A特征值为$\lambda$的特征向量。

  由于A为实对称方阵$\Rightarrow A^H=A$
  \begin{equation}
    \begin{split}
      \Rightarrow \bar{\lambda}X^HX& =(\bar{\lambda}X^H)X=(X^HA^H)X=X^HA^HX=X^HAX \\
      & = X^H(AX)=X^H(\lambda X)=\lambda X^HX
    \end{split}
  \end{equation}
  得$\lambda \in \mathbb{R}$
\end{proof}

\begin{remark}
  $A^H$代表$A$的共轭转置。
\end{remark}


\begin{proof}
  归纳得，$n=1$时显然成立。

  假设$n=k\in \mathbb{N}^*$时结论成立，考虑$k+1$阶实对称矩阵$A_{k+1}$：
  根据引理，实对称矩阵$A_{k+1}$存在k+1个实特征值（可能相等），对应k+1个$R^{k+1}$中的特征向量。

  设其中特征值为$\lambda_1$的单位特征向量为$X_1$.
  
  将$X_1$扩充为$R^{k+1}$中的以组标准正交基，令$T_0=(X_1,X_2,\cdots,X_{k+1})\Rightarrow T_0$为正交阵。

  则\begin{equation}\label{2}
  \begin{split}
    T_0^{-1}AT_0 & =T_0^TAT_0=\left(\begin{array}{c}
      X_1^T \\
      X_2^T \\
      \vdots \\
      X_{k+1}^T
    \end{array} \right)A\left(X_1\ X_2\ \cdots \ X_{k+1}\right)\\
    & =\left(\begin{array}{cccc}
      X_1^TAX_1 & X_1^TAX_2 & \cdots & X_1^TAX_{k+1} \\
      X_2^TAX_1 & X_2^TAX_2 & \cdots & X_2^TAX_{k+1} \\
      \vdots & \vdots & \ddots & \vdots \\
      X_{k+1}^TAX_1 & X_{k+1}^TAX_2 & \cdots & X_{k+1}^TAX_{k+1}
    \end{array}\right)
  \end{split}
  \end{equation}
  注意到
  \begin{equation}
    X_1^TAX_1=X_1^T(AX_1)=X_1^T(\lambda X_1)=\lambda X_1^TX_1=\lambda
  \end{equation}
  \begin{equation}\label{}
    X_j^TAX_i=\lambda X_j^TX_i=0(i\neq j)
  \end{equation}
  从而
  \begin{equation}\label{}
    \begin{split}
      \left(\begin{array}{cccc}
        X_1^TAX_1 & X_1^TAX_2 & \cdots & X_1^TAX_{k+1} \\
        X_2^TAX_1 & X_2^TAX_2 & \cdots & X_2^TAX_{k+1} \\
        \vdots & \vdots & \ddots & \vdots \\
        X_{k+1}^TAX_1 & X_{k+1}^TAX_2 & \cdots & X_{k+1}^TAX_{k+1}
      \end{array}\right) & =\left( \begin{array}{cccc}
        \lambda_1 & 0 & \cdots & 0 \\
        0 & X_2^TAX_2 & \cdots & X_2^TAX_{k+1} \\
        \vdots & \vdots & \ddots & \vdots \\
        0 & X_{k+1}^TAX_2 & \cdots & X_{k+1}^TAX_{k+1}
      \end{array}\right) \\
      & =\left( \begin{array}{cccc}
        \lambda_1 & 0 & \cdots & 0 \\
        0 &  &  &  \\
        \vdots &  & A_k &  \\
        0 &  &  &
      \end{array}\right)
    \end{split}
  \end{equation}
  其中$A_k$为k阶对称实矩阵。由假设知，$\exists T_1,s.t.T_1^{-1}A_kT_1 =\begin{pmatrix}
  \lambda_1 &   &   &   \\
  & \lambda_2 &   &   \\
  &   & \ddots  &   \\
  &   &   & \lambda_{k+1}
  \end{pmatrix} $

  $\Rightarrow A_k=T_1\begin{pmatrix}
  \lambda_1 &   &   &   \\
  & \lambda_2 &   &   \\
  &   & \ddots  &   \\
  &   &   & \lambda_{k+1}
  \end{pmatrix}T_1^{-1}$

  所以原式有
  \begin{equation}\label{}
    \begin{split}
      \left( \begin{array}{cccc}
        \lambda_1 & 0 & \cdots & 0 \\
        0 &  &  &  \\
        \vdots &  & A_k &  \\
        0 &  &  &
      \end{array}\right) & \left( \begin{array}{cccc}
        \lambda_1 & 0 & \cdots & 0 \\
        0 &  &  &  \\
        \vdots &  & T_1\left(\begin{array}{ccc}
          \lambda_2 &  &  \\
          & \ddots &  \\
          &  & \lambda_{k+1}
        \end{array} \right) T_1^{-1}&  \\
        0 &  &  &
      \end{array}\right) \\
      & \left( \begin{array}{cc}
        1 & O \\
        O & T_1
      \end{array}\right)\left(\begin{array}{cccc}
        \lambda_1& & & \\
        & \lambda_2 &  &  \\
        & & \cdots &  \\
        & & & \lambda_{k+1}
      \end{array} \right)\left( \begin{array}{cc}
        1 & O \\
        O & T_1^{-1}
      \end{array}\right)
    \end{split}
  \end{equation}
  令$T=T_0\left(\begin{array}{cc}
    1 & O \\
    O & T_1
  \end{array}\right)$,即有$A_{k+1}=T^{-1}diag{\lambda_1,\lambda2,\cdots,\lambda_{k+1}}T$
  又有$$T^TT=\left(\begin{array}{cc}
    1 & O \\
    O & T_1^T
  \end{array}\right)T_0^TT_0\left(\begin{array}{cc}
    1 & O \\
    O & T_1
  \end{array}\right)=I_{k+1}$$
  综上命题成立。
\end{proof}


\end{document}
